Evaluate the expression \[ (a^2+b)^2 - (a^2-b)^2, \]if $a=4$ and $b=1$.
Explanation: Perhaps the fastest way is to use the difference of squares factorization: \begin{align*}
(a^2 + b)^2 - (a^2 - b)^2 &= \bigl[ (a^2 + b) + (a^2 - b) \bigr] \cdot
\bigl[ (a^2 + b) - (a^2 - b) \bigr] \\
&= ( a^2 + b + a^2 - b) \cdot (a^2 + b - a^2 +b ) \\
&= (2 a^2 ) \cdot (2 b) \\
&= 4 a^2 b. \end{align*}Since $a= 4$ and $b=1$, this last expression is equal to \[ 4 \cdot 4^2 \cdot 1 = 4 \cdot 16 = \boxed{64}, \]so that is our answer.

We could also plug in the values of $a$ and $b$ right away and then expand.  We then get \begin{align*}
(a^2 + b)^2 - (a^2 - b)^2 &= (4^2 + 1)^2 - (4^2 -1)^2 \\
&= (16 + 1)^2 - (16- 1)^2 \\
&= 17^2 - 15^2 . \end{align*}Now, $17^2 = 289$, and $15^2 = 225$, so our answer is then \[ 289 - 225 = 89 -25 = 64, \]as before.